Gulp 4 watch all sass files, compile one - npm

Question

My file structure is:
sass
-main.sass
-variables.sass
-containers.sass
-buttons.sass
-helpers.sass
css
-main.css
In my main.sass file, I'm importing all the other sass files:
#import "variables";
#import "containers";
#import "buttons";
//etc...
In gulp I want to watch any changed scss file and compile only main.sass into main.css.
This is what I have:
var gulp = require('gulp');
var sass = require('gulp-sass');
function styles() {
return gulp.src('sass/main.sass', {
sourcemaps: true
})
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('css/'));
}
function watch() {
gulp.watch('sass/**/*.sass', styles);
}
var build = gulp.parallel(styles, watch);
gulp.task(build);
gulp.task('default', build);
The files are being watched but when I update one, the main.css file is not updated.

Answer 1

I believe one of your problems could be that you define a watching task via gulp.watch(...) but its reference is lost and so it never gets called. However, I don't really understand what you try to achieve with the call to parallel(). I'm currently working on a similar use case with a build script like the following:
var gulp = require('gulp');
function styles() {
return gulp.src('...')
.pipe(...)
.pipe(gulp.dest('...'));
}
exports.build = gulp.watch('...', styles);
This should be a minimal working example, just call gulp build.